Go to the video lesson for this slide deck: h2p://edrolo.com.au/subjects/physics/hsc-­‐physics/space-­‐part-­‐2/escape-­‐velocity/lesson/ HSC Physics – Core 9.2 Space!
Part 2: Launching into orbit!
Overview of Part 2:!
2.1 Projectile motion (PM)!
2.2 Escape velocity – getting into orbit!
2.3 Accelerations and g-forces during launch!
2.4 Uniform circular motion (UCM)!
2.5 Types of orbits!
2.6 Kepler’s Law of periods!
2.7 Total orbital energy of a satellite!
2.8 The sling-shot effect (gravity-assist)!
2.9 Orbital decay!
2.10 Re-entry!
!
HSC Physics – Core 9.2 Space!
Part 2: Launching into orbit!
!2.2: Escape velocity!
Overview of Section 2.2:!
2.2.1 Escape velocity formula!
2.2.2 Velocity assist from Earth’s axial rotation!
2.2.3 Velocity assist from Earth’s orbital motion !
!
!
Escape velocity!
Syllabus Module 9.2 dot point 2B4 Outline Newton’s concept of escape velocity!
!
Syllabus Module 9.2 dot point 2B6 Discuss the effect of the Earth‘s orbital motion and its
rotational motion on the launch of a rocket!
!
Escape velocity Formula!
Escape velocity, ues !
= the initial velocity (u) at the surface of the Earth that any
object must be given for it to just escape the Earth’s
gravitational attraction.!
!
!“escape Earth’s gravitational attraction” means
!
!the object will not fall back to the Earth or orbit the Earth.!
!
!“just escape” means that just as the object reaches an infinite
distance from the Earth its final velocity = 0
!!
By conservation of energy
Ek ( at surface ) + E p ( at surface ) = E p ( at infinte distance )
⎛ 1
⎞ ⎛ G mrocket mEarth ⎞
2
⎟ = 0
m
u
⎜ 2 rocket escape ⎟ + ⎜⎜ −
⎟
r
⎝
⎠ ⎝
E
⎠
So
uescape =
2GmEarth
rEeart
… Exercise 1 !
Escape velocity – Exercise 1
!
Calculate the escape velocity from Earth.!
!
Data:!
Mass of Earth
!= 5.98 x 1024 kg!
Radius of Earth
!= 6380 km!
Universal Gravitational Constant G = 6.67 x 10-11 N kg-2 m!
… Answer!
Escape velocity – Exercise 1: Answer !
Calculate the escape velocity from Earth.!
Data:!
Mass of Earth
!= 5.98 x 1024 kg!
Radius of Earth
!= 6380 km!
Universal Gravitational Constant G = 6.67 x 10-11 N kg-2 m2!
To find ues from the Earth
use
ues =
=
2GmEarth
RE
2(6.67 x10-11)(5.98 x1024 )
(6.380 x106 )
= 11.182 × 103 m s -1
≈ 11.2 km s -1
Escape velocity – Exercise 2!
When a spacecraft is given its escape velocity, !
(a)  Does the direction in which it is launched matter!
(b)  Does the path it follows matter? !
… Answer!
Escape Velocity – Exercise 2: Answer (a)!
When a spacecraft is given its escape velocity,!
(a) does the direction in which it is launched matter?!
Direction DOES matter: !
To save fuel it is best to launch the vehicle towards the east to
make use of the Earth’s axial rotation towards the East. !
… Answer (b)!
Escape Velocity – Exercise 2: Answer (b)!
When a spacecraft is given its escape velocity, !
(b) Does the path it follows matter?!
Escape velocity is calculated by equating the kinetic energy at launch
on the surface (r = rsurface) with the gain in gravitational potential
energy the body would have at infinity (r = !)!
(Ek )surface = ΔEp
= (Ep ) − (Ep )
∞
surface
⎛ Gmplanetmspacecraft ⎞
= 0 − ⎜ −
⎟⎟
⎜
r
planet
⎝
⎠
Gmplanetmspacecraft
=
rplanet
Next …!
Launch velocity-assist from Earth’s axial-rotation – !
So the kinetic energy needed at launch to escape the planet does
not depend on the path the spacecraft follows. !
It depends only on the radius of the planet it is escaping!
Launch velocity-assist from Earth’s axial-rotation !
The size of the tangen&al speed of the Earth from east to west is different at different laCtudes: vθ = v equator cos θθ
This velocity can be used to assist a rocket at launch but the vehicle must be launched towards the East. but
v
r
ω = constant
so
ωθ = ωEquator
so
vθ vEquator
=
rθ
rE
Pr oof
ω=
vθ = vEquator
rE = rEquator
rθ
rE
vθ = vEquator cos θ
… Example 1!
Launch velocity-assist from Earth’s axial-rotation – Example 1 !
What is the tangential velocity of the Earth’s surface at the
Equator.!
!
Earth Data:!
Period of rotation = 23 hours 56 minutes 4 seconds!
Radius of Earth = 6371 km. Hint: !
If the Earth rotates uniformly on its axis, then: v Equator
2π rEarth
=
TEarth
… Answer!
Launch velocity-assist from Earth’s axial-rotation – Example 1: Answer !
Find tangential velocity of the Earth’s surface at the Equator Earth data
Period of rotation of Earth, TRotation = 23 h 56 min
Radius of Earth,
rE = 6371 km
To find vEquator
Use
vEquator =
2π rEarth
TEarth
2π × 6371× 103 m
=
TEarth
vEquator
2π 6371× 103 m
=
86164 s
= 464.6 m s -1
Convert units of TRotation to seconds
TRotation = 23 h 56 min
= 23 × 3600 s + 56 × 60 s
= 86164 s
… Exercise 3 !
Launch velocity-assist from Earth’s axial-rotation – Exercise 3:!Answer!
What is the tangential velocity of the Earth’s surface at Sydney
(latitude 340 S). !
Data
v equator = 464.6 m s-1 (from Example 1)
θsydney = 34o South
To find v Sydney
Use
vθ = v equator cos θ
v 34o = 464.6 × cos 34o
∴ v Sydney = 385.2 m s-1
… Exercise 4 !
Launch velocity-assist from
Earth’s axial rotation – Exercise 4!
Australia had a rocket launch site at Woomera in South
Australia (latitude 30o S) from the 1950s to 1970s. !
There have been plans to build a new launch site on Cape
York (latitude 120 S). !
Suppose you want to launch a rocket so that it will escape
the Earth.!
!
Which location (Woomera or Cape York) would be the best
site to locate the launch site? Explain why.!
… Answer!
Launch velocity-assist from
Earth’s axial rotation – Exercise 4: Answer!
Which location (Woomera or Cape York) would be the best site to
locate the launch site? Why?!
The tangential velocity of the Earth’s surface at these two latitudes is
calculated below. !
Earth's tangential velocity at Woomera
Earth's tangential velocity at Cape York
Use
Use
v latitude,θ = vequator cosθ
v latitude,θ = vequator cosθ
Latitude θCape York = 12o
Latitude θ Woomera = 30o
vequator = 464.6 m s-1
vWoomera = 464.6 x cos 300
≈ 402.4 m s-1
vequator = 464.6 m s-1
vCape York = 464.6 x cos120
≈ 454.4 m s-1
The best location will be the one with the greatest assistance from the
Earth’s rotational motion at launch.!
This occurs for the site with the smallest latitude (i.e. the site nearer the
equator) = Cape York. !Next …!
Launch velocity-assist from Earth’s orbital motion!
Launch velocity-assist from
Earth’s orbital-motion about the Sun!
Once a vehicle is in orbit, the Earth’s orbital velocity
about the Sun is added to the vehicle’s orbital
velocity about the Earth.!
The Earth’s orbital velocity
about the sun VES is about
30 km s-1 in an
anti-clockwise direction!
r
vrocket
relative to Sun
r
r
= vrocket's orbital velocity + vEarth's orbital velocity
about the Earth
r
r
r
vRS = vRE + vES
about the Sun
… Exercise 5 !
Launch velocity-assist from
Earth’s orbital-motion about the Sun – Exercise 5!
Explain how you could take advantage of the Earth’s orbital
motion about the Sun to reach: !
(a) an outer planet – one further from the Sun than the Earth!
(b) an inner planet – one closer to the Sun than the Earth. … Answer!
Launch velocity-assist from Answer (a)
Earth’s orbital-motion about the Sun – Exercise 5!
Explain how you could take advantage of the Earth’s orbital
motion about the Sun to reach:!
(a) An outer planet – one further from the Sun than the Earth!
Spacecraft heading to outer planets are not launched until the
direction of the Earth in its orbit around the Sun corresponds
with the desired direction. !
The spacecraft can then be launched up to a low Earth orbit,
before firing its rockets again to accelerate ahead of the Earth!
The velocity boost received from the Earth’s orbital motion in
this case is approximately 30 km s-1. … Answer (b)!
Launch velocity-assist from Answer (b)
Earth’s orbital-motion about the Sun – Exercise 5!
Explain how you could take advantage of the Earth’s orbital motion
about the Sun to reach:!
(b) an inner planet – one closer to the Sun than the Earth. A spacecraft traveling from Earth to an inner planet will be naturally
accelerated towards the planet by the Sun’s gravity. !
To assist the spacecraft, it should be:!
•  launched in the direction opposite to the orbital motion of the
Earth about the Sun (i.e. in a clockwise direction): this will
reduce its total orbital energy causing it to descent to a lower
orbital radius about the Sun !E = − 1 GmM ( if E becomes more negative, r ↓)
2 r
T
!
T
o
o
•  decelerated (using its rocket engines) until it achieves a sun-orbit
with a perihelion equal to the orbit of the inner planet. !
So, the spacecraft will continue to move in the same direction as
Earth, only more quickly. … next video: Accelerations and g-forces!